It's school time!

Let $b,p$ and $n$ be the cost of one bag, one pen and one notebook respectively.

$2b+4p=30$ $\implies$ $4p=30-2b$ $\implies$ $p=\dfrac{15-b}{2}$ $\color{#D61F06}(1)$

$5p+4n=29$ $\implies$ $5p=29-4n$ $\implies$ $p=\dfrac{29-4n}{5}$ $\color{#D61F06}(2)$

$3n+3b=37.5$ $\color{#D61F06}(3)$

Equate $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ .

$\dfrac{15-b}{2}=\dfrac{29-4n}{5}$

$75-5b=58-8n$

$-8n+5b=17$ $\color{#D61F06}(4)$

$(3\times$ $\color{#D61F06}(4)$ $)+(8\times$ $\color{#D61F06}(3)$ $)$ , we get

$39b=351$

$b=9$

It follows that $p=\dfrac{15-9}{2}=3$ .

Solving for $n$ , we have

$3n+3(9)=37.5$ $\implies$ $3n=10.5$ $\implies$ $n=3.5$

Finally,

$b+p+n=9+3+3.5=$ $\boxed{15.5}$